Answer
$$0$$
Work Step by Step
$$\eqalign{
& {\sin ^{ - 1}}x + {\sin ^{ - 1}}\left( { - x} \right) \cr
& {\text{Recall that si}}{{\text{n}}^{ - 1}}\theta {\text{ is an odd function, then }}{\sin ^{ - 1}}\left( { - x} \right) = - {\sin ^{ - 1}}x \cr
& {\text{then,}} \cr
& {\sin ^{ - 1}}x - {\sin ^{ - 1}}x \cr
& 0 \cr} $$